\( \sum _ { i = 4 } ^ { m } 8 ( \frac { 2 } { 3 } ) ^ { i - 5 } = \frac { 2660 } { 81 } \)

To solve the equation \( \sum _ { i = 4 } ^ { m } 8 \left( \frac { 2 } { 3 } \right) ^ { i - 5 } = \frac { 2660 } { 81 } \), we first recognize that this series is a geometric series. The first term of the series when \( i = 4 \) is \( 8 \left( \frac{2}{3} \right)^{-1} = 12 \) and the common ratio is \( \frac{2}{3} \). This series can be expressed as: \[ 12 + 8 \left( \frac{2}{3} \right)^{0} + 8 \left( \frac{2}{3} \right)^{1} + \dots \] The sum of a geometric series can be determined using the formula: \[ S_n = a \frac{1 - r^n}{1 - r} \] where \( a \) is the first term, \( r \) is the common ratio, and \( n \) is the number of terms. In this case, the first term is \( 12 \), the common ratio is \( \frac{2}{3} \), and the number of terms \( n = m - 4 + 1 = m - 3 \). Thus, we can write the sum as: \[ 10 \cdot \frac{1 - \left( \frac{2}{3} \right)^{m-3}}{1 - \frac{2}{3}} = 12 \cdot \frac{1 - \left( \frac{2}{3} \right)^{m-3}}{\frac{1}{3}} = 36 \left( 1 - \left( \frac{2}{3} \right)^{m-3} \right) \] Now set this equal to \( \frac{2660}{81} \): \[ 36 \left( 1 - \left( \frac{2}{3} \right)^{m-3} \right) = \frac{2660}{81} \] To isolate the term with \( m \), divide both sides by \( 36 \): \[ 1 - \left( \frac{2}{3} \right)^{m-3} = \frac{2660}{2916} \] Subtract 1 from both sides: \[ -\left( \frac{2}{3} \right)^{m-3} = \frac{2660 - 2916}{2916} = \frac{-256}{2916} \] Thus, \[ \left( \frac{2}{3} \right)^{m-3} = \frac{256}{2916} \] Now, simplify \( \frac{256}{2916} \) and solve for \( m \). Note that \( 256 = 2^8 \) and \( 2916 = 2 \times 3^6 = 2 \times 729 \) simplifying in exponent terms we obtain: Since \( \frac{256}{2916} = \left( \frac{2}{3} \right)^{m-3} \) By taking log base \( \frac{2}{3} \): \[ m - 3 = \log_{\frac{2}{3}} \left( \frac{256}{2916} \right) \] And thereby: \[ m = 3 + \log_{\frac{2}{3}} \left( \frac{256}{2916} \right) \] Calculate it using the logarithm if necessary, and determine the value of \( m \).